Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 74 / 3

Let ${\rho }_0$ be the reference mass density, $\nu$ the kinematic viscosity, $\kappa$ the thermal diffusivity, $\eta$ the magnetic diffusivity, and ${\alpha }_T$ the coefficient of thermal expansion. The conductive state of Rayleigh-Bénard convection is $T_b=T_0+\Delta T(1-z/d)$. For the Boussinesq approximation, density variation is retained only in buoyancy. Linearizing about this state and the imposed magnetic field $B_0\hat{\mathbf{y}}$ givesHere $w=u_z$ and $p_{\ast }=p^{\prime }+B_0{b}_y/{\mu }_0$ is the perturbation of magnetohydrodynamic total pressure. The gradient part of the Lorentz force has been absorbed into $p_{\ast }$; the remaining term is magnetic tension. The sign in the temperature equation follows from $d{T}_b/dz=-\Delta T/d$.

Use the thermal-diffusion scaling of a convection layer: length $d$, time $d^2/\kappa$, velocity $\kappa /d$, temperature perturbation $\Delta T$, field perturbation $B_0$, and total-pressure perturbation ${\rho }_0\nu \kappa /d^2$. Dropping dimensionless-variable decorations giveswithThese are respectively the Rayleigh number, Chandrasekhar number, Prandtl number, and magnetic-to-thermal diffusivity ratio. In particular, the magnetic coefficient is $\zeta Q$, not $Q$ alone, in these thermal-time units. At $z=0,1$, the boundary conditions are $\theta =w=b_z=0$, ${\partial }_z{u}_x={\partial }_z{u}_y=0$, and ${\partial }_z{b}_x={\partial }_z{b}_y=0$.

Put $a_h^2=k^2+m^2>0$ and $\Lambda =a_h^2+{\pi }^2$. The plane-layer poloidal-toroidal decomposition supplies the assumed poloidal fields asSince $w=-{\Delta }_h\varphi$ and $b_z=-{\Delta }_h\chi$, potentials proportional to $e^{st+ikx+imy}\sin \pi z$ have the required vertical components. Their horizontal components are proportional to $\cos \pi z$, satisfying the stress-free boundary conditions and the horizontal-field boundary conditions. The double-curl representation automatically enforces incompressibility and the solenoidal magnetic-field constraint.

Let $W,\Theta ,C$ be the amplitudes of $w,\theta ,b_z$. The temperature and resistive induction equations immediately giveEliminate the total-pressure perturbation by applying $\Delta$ to the vertical momentum equation and subtracting ${\partial }_z$ times its divergence. The resulting vertical equation isIn particular, $imC=-m^{2}W/(s+\zeta \Lambda )$ shows explicitly that magnetic tension opposes the displacement.

Taking the determinant of these three amplitude equations gives the horizontal-field magnetoconvection dispersion relation:This cubic polynomial includes all roots of the assumed poloidal coupled system; using its determinant form avoids discarding possible roots by dividing by $s+\Lambda$ or $s+\zeta \Lambda$. As a check, $Q{m}^2=0$ factors off magnetic diffusion and leaves the standard stress-free Rayleigh-Bénard convection growth equation.

Set $s=0$ in the horizontal-field magnetoconvection dispersion relation. With positive $\sigma ,\zeta$ and nonzero $a_h$, cancellation gives the stationary neutral curveAt fixed $a_h^2$, the Chandrasekhar number contributes a nonnegative penalty proportional to $m^2$. It can be eliminated by choosing $m=0$. It remains to minimize $f(x)=(x+{\pi }^2{)}^3/x$ for $x=k^2>0$:The function diverges at both endpoints and has its unique minimum at $x={\pi }^2/2$. Thus the magnetic-field-aligned rolls evade steady magnetic inhibition:independently of $Q$. If higher vertical sine modes are included, replacing $\pi$ by $n\pi$ gives minimum $27n^4{\pi }^4/4$, so the displayed first vertical mode also gives the smallest steady threshold.

Physically, these convection rolls have axes along the imposed magnetic field. Their perturbations are independent of $y$, so ${\partial }_y\mathbf{u}=0$ in the resistive induction equation and no growing magnetic perturbation is generated. The flow does not bend the field; consequently magnetic tension cannot inhibit these rolls, however large the field strength. This conclusion concerns the requested steady disturbances.

For disturbances varying only along the imposed magnetic field, set $k=0$ and $x=m^2>0$. The steady horizontal-field magnetoconvection dispersion relation becomesExpand it in powers of $x$:Its derivative and second derivative areBecause $R$ diverges as $x\to 0$ and $x\to \infty$, the unique minimum is determined byFor large Chandrasekhar number, this equation first gives $x\sim {\pi }^3{Q}^{-1/2}$ and hence the requested strong-field steady magnetoconvection varying along the field selection law:

For the threshold through order unity, the leading $Qx+{\pi }^6/x$ contribution at its optimum is $2{\pi }^3\sqrt{Q}$. More precisely, the stationary equation givesThe correction to $Qx+{\pi }^6/x$ has no first-order contribution because that expression is already stationary at its leading minimizer. Also $3{\pi }^{2}x=O(Q^{-1/2})$ and $x^2=O(Q^{-1})$. ThereforeThe constant term $3{\pi }^4$ must be retained. This restricted family bends magnetic field lines, so its large stationary threshold is compatible with the unrestricted field-aligned minimum in the preceding part.